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testtable1 = {1,1,1,1,2,2,2,2,3,3,3,3,5,5,5}
I want to create another table who one element of each in testtable1 --->
testtable2 = {1,2,3,5}
I think there is a kind of table.something? Like table.sort?
Ty for helping
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Scripts How to take one of each in a table?
How to take one of each in a table?
22 repliestesttable1 = {1,1,1,1,2,2,2,2,3,3,3,3,5,5,5}
I want to create another table who one element of each in testtable1 --->
testtable2 = {1,2,3,5}
I think there is a kind of table.something? Like table.sort?
Ty for helping
Flacko, The Ultimate Lua Challenge has written1
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function unique(t)
local r = {}
local i = {}
for k,v in pairs(t) do
if not i[v] then
r[#r+1] = v
i[v] = 1
end
end
return r
end
a = unique({1,4,5,2,2,3,4})
for k,v in pairs(a) do print(v) end
How do i remove a random variable from testtable1 then?
edited 2×, last 09.05.17 02:48:13 pm
table.remove()many times.
1
table.remove( testtable1, math.random(1, #testtable1) )
math.random()picks randomly a value from the table and then it gets removed.
edited 2×, last 09.05.17 06:47:31 pm
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function randomt(table, number)
local t = {}
for k, v in pairs(table) do
if v == number then
table.insert(t, k)
end
end
if #t > 0 then
table.remove(table, math.random(#t))
end
end
Is that possible to sort a table without using table.sort. And when you sort your table it sort in order to the number who comes first.
For example, i have a table.
Table = {3,1,3,1,3,2,1,2}
It sorts to
Table = {3,3,3,1,1,1,2,2}
I don't want to use Table.sort cause it will make the table -->
Table = {1,1,1,2,2,3,3,3}
TestTable = {3,4,3,2,1,4,2,1,3}
I mean i want to sort a table in a order of the first number who comes first and then second one, and the third one etc....
The first number who appear in the table is "3" right. Then collect all the "3" in the table who comes first.
then the table should look like this
TestTable = {3,3,3,4,2,1,4,2,1}
then the second number is "4" to collect. then collect all the number "4" and the table should looks like
TestTable = {3,3,3,4,4,2,1,2,1}
then collect all number 2
TestTable = {3,3,3,4,4,2,2,1,1} then its done
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function differents(table)
local d = 0
local c
for k, v in pairs(table) do
if not c or c ~= v then
d = d + 1
c = v
end
end
return d
end
function tableif(t,s)
for k, v in pairs(t) do
if v == s then
return true
end
end
return false
end
function tablesort(table)
local t = {}
local c = false
for i = 1, differents(table) do
for k, v in pairs(table) do
if not c and not tableif(t,v) then
c = v
end
if v == c then
table.insert(t,v)
end
end
c = false
end
return t
end
differentstells me how many different things in table.
tableiftells me if selected number is in table. And finally,
tablesortis what you are looking for.
I made it too quickly, it can be done with using fewer lines and might even not work. So test it, I'll always be here to help.
edited 2×, last 09.05.17 09:07:25 pm
tablesort(TestTable)
Line 10
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function tablesort(table)
local t = {}
local c = false
for i = 1, differents(table) do
for k, v in pairs(table) do
if not c and not tableif(t,v) then
c = v
end
if v == c then
---->table.insert(t,v)<-----
end
end
c = false
end
return t
end
edited 1×, last 09.05.17 08:50:32 pm
EDIT: Oh fuck, use this one. I used "table" as parameter name, shiet.
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function tablesort(tables)
local t = {}
local c = false
for i = 1, differents(tables) do
for k, v in pairs(tables) do
if not c and not tableif(t,v) then
c = v
end
if v == c then
table.insert(t,v)
end
end
c = false
end
return t
end
I just went on a little problem on the sortation table function.
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function differents(table)
local d = 0
local c
for k, v in pairs(table) do
if not c or c ~= v then
d = d + 1
c = v
end
end
return d
end
function tableif(t,s)
for k, v in pairs(t) do
if v == s then
return true
end
end
return false
end
function tablesort(tables)
local t = {}
local c = false
for i = 1, differents(tables) do
for k, v in pairs(tables) do
if not c and not tableif(t,v) then
c = v
end
if v == c then
table.insert(t,v)
end
end
c = false
end
return t
end
function unique(t)
local r = {}
local i = {}
for k,v in pairs(t) do
if not i[v] then
r[#r+1] = v
i[v] = 1
end
end
return r
end
When i am sorting a table with over 200 units in it. It will freeze the server for 1 second and then works again. Is there possible to avoid the big loop and avoid the freeze time for the server when i have over 200 units in a table?
When the table are over 200 variables, it starts lagging cause of the big loop from tablesort(tables). Is that possible to do it without a big loop to avoid the lagg?
edited 6×, last 09.10.17 04:37:03 pm
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TestTable = {3,4,3,6,2,1,6,4,2,6,6,1,3,6}
print(table.concat(TestTable, ", "))
function greetSort(tbl)
-- the smaller the value - the closer to the beginning of the sorted table a value must be
-- small values = first
-- large values = last
local nextValAt = {}
local itemsMoved, steps = 0, 0
for i = 1, #tbl do
steps = steps + 1
local item = tbl[i]
if not nextValAt[item] then
nextValAt[item] = i+1
--print("i=".. i, "New unknown value: ".. item, "New Rank: ".. i+1)
-- don't try to replace the currently existing same item
-- e.g: 5,5 - don't start the copying process of the latter 5
elseif item == tbl[nextValAt[item]] then
nextValAt[item] = nextValAt[item] + 1
--print("Item ".. item .." is in the correct pos. Increasing value of ".. item .." by 1 -> ".. nextValAt[item])
else
--print("i=".. i, " known value: ".. item, "Inserting as #".. nextValAt[item], "Between ".. (tbl[nextValAt[item]-1] or "@") .." and ".. (tbl[nextValAt[item]] or "@"))
table.insert(tbl, nextValAt[item], item)
itemsMoved = itemsMoved + 1
repeat
nextValAt[item] = nextValAt[item] + 1
--print("Increasing value of ".. item .." by 1 -> ".. nextValAt[item])
if nextValAt[item] > i then break end
item = tbl[nextValAt[item]]
steps = steps + 1
until item == nil
table.remove(tbl, i+1)
end
end
print("steps: ".. steps, "itemsMoved: ".. itemsMoved)
end
greetSort(TestTable)
print(table.concat(TestTable, ", "))
It goes through the elements of the table and once it discovers an unknown element, this element receives a "rank" in the table (this rank is actually just the position in the table). If the same elements are discovered again in the table, it will be moved to the right of the chain of equal elements: 5,4,3,5 > 5,5,4,3
Performance: waaaay better than Masea's approach (@
Masea: you iterate the table as many times as many different types of elements exist if I read it correctly).It's good on tables with many similar elements and gets a little worse if there are a lot of different elements.
I do realise there're better ways to do this, but I just had fun trying to do that within a single pass/single run of the for-loop. I think it would be faster if you create a separate table for each element and then glue them together into a big table.
You can benchmark them yourself, just don't forget to save Masea's final script in masea-sort.lua:
Destructive function (changes the input table) using table.sort
edited 2×, last 11.10.17 11:53:30 pm
